It sounds beautiful and I was wondering if there's any nice proof. A friend tells me there's a more general theory about some so-called Hilbert functions which justify this, but I'm interested in something easier!

Do you mean that for each $\alpha$ there exists $A_\alpha$ such that $|f_1(x) - f_1(y)|\leq A_\alpha|x - y|^\alpha$? Or do you mean that there exists an $\alpha\in(0,1)$ and $A_\alpha > 0$ such that the above inequality holds? (I think it's the latter, since the former would imply that $f_1$ is Lipschitz). If it is the latter, then you can simply say that $f_1$ is $\alpha$-Hölder continuous; see here: en.wikipedia.org/wiki/H%C3%B6lder_condition.
–
WilliamMar 28 '12 at 5:04

@WNY: Why would the former imply that $f_1$ is Lipschitz?
–
DidMar 28 '12 at 5:20

1

@Didier: The OP didn't introduce $\alpha$ and $A_\alpha$ in the original post, and I was making guesses; it was only after I posted my comment that the OP edited the original question. Sorry for the confusion.
–
WilliamMar 28 '12 at 6:11

Lets divide the sum into two:
$$
\sum_{n=0}^{N}{2^{-n} e^{2\pi i 2^{n} x} }
+\sum_{n=N+1}^{\infty}{2^{-n} e^{2\pi i 2^{n} x} }=S_1(x)+S_2(x).
$$
The difference of the first sum can be estimated by the mean value theorem:
$$
|\Delta S_1(x)|\le \sum_{n=0}^{N}{2^{-n} (2\pi 2^{n} |\Delta x|)} =2\pi (N+1)|\Delta x|,
$$
and the second are marjorized by the sum of an infinite geometric progression:
$$
|\Delta S_2(x)|\le \sum_{n=N+1}^{\infty}{2^{-n}}=2^{-N}.
$$
Now for given $\Delta x$ one can choose $N$ s.t. both summands satisfy the required estimate.